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2016 Galois Contest

Wednesday, April 13, 2016(in North America and South America)

Thursday, April 14, 2016(outside of North America and South America)

Solutions

©2016 University of Waterloo

2016 Galois Contest Solutions Page 2

1. (a) The first bucket contains 7 red discs.Each bucket after the first contains 3 more red discs than the previous bucket.Thus, the second bucket contains 7+3 = 10 red discs, the third bucket contains 10+3 = 13red discs, and the fourth bucket contains 13 + 3 = 16 red discs.

(b) Solution 1Let b represent the number of buckets after the first.Since the first bucket contains 17 green discs and each bucket after the first contains 1more green disc than the previous bucket, then there are 17 + b green discs inside thebucket b after the first.Since the first bucket contains 7 red discs and each bucket after the first contains 3 morered discs than the previous bucket, then there are 7 + 3b red discs inside the bucket b afterthe first.The number of green discs in a bucket is equal to the number of red discs inside the samebucket when 17 + b = 7 + 3b or when 2b = 10, and so when b = 5.Thus, there are an equal number of red discs and green discs in the bucket 5 after the firstbucket, which is the 6th bucket.

Solution 2Using the fact that the first bucket contains 17 green discs and 7 red discs, and each bucketafter the first contains 1 more green disc and 3 more red discs than the previous bucket,then we may summarize the number of green and red discs inside each bucket.

Bucket Number Number of green discs Number of red discs1 17 72 18 103 19 134 20 165 21 196 22 22

Therefore, the 6th bucket contains an equal number of red discs and green discs.(Note that since the number of red discs is increasing by 3 each bucket and the numberof green discs is increasing by 1 each bucket, then this is the only bucket in which thenumber of red discs and green discs will be equal.)

Solution 3In the first bucket, there are 17 green discs and 7 red discs. Each bucket after the firstcontains 1 more green disc and 3 more red discs than the previous bucket.Since there are 2 more red discs than green being put in, then the difference between thenumbers of green and red discs will decrease by 2 for each bucket after the first.Since the original difference is 17− 7 = 10, then it takes 10÷ 2 = 5 more buckets to arriveat a bucket where the numbers of green and red discs will be equal.Therefore, the 6th bucket contains an equal number of red discs and green discs.

(c) As in part (b), there are 17 + b green discs and 7 + 3b red discs inside the bucket b afterthe first.The number of red discs in a bucket is equal to twice the number of green discs inside thesame bucket when 7 + 3b = 2(17 + b) or when 7 + 3b = 34 + 2b, and so when b = 27.In the 27th bucket after the first (the 28th bucket), there are 17 + 27 = 44 green discs and7 + 3(27) = 88 red discs. (We note that 88 is indeed twice 44.)The total number of discs in this bucket is 44 + 88 = 132.

2016 Galois Contest Solutions Page 3

2. (a) A plate with 36 shaded squares has 10 shaded squares alongeach side of the plate, as shown.We can see this from the diagram, or by considering a platewith s squares along each side.In this case, we can count s squares on the top edge and ssquares on the bottom edge, plus s− 2 new squares on eachof the left edge and the right edge. (The 2 corner squares oneach of these edges are already counted.)This means that there are 2s + 2(s − 2) = 4s − 4 squaresalong the edges.Here, we want 4s− 4 = 36 or 4s = 40, and so s = 10.There are 10 squares along each side of the plate and theside length of the square plate is 60 cm, thus the side lengthof each of the shaded squares is 60

10= 6 cm.

60 cm

60 cm

(b) Since the plate is a square, and there are an equal number of identicalshaded squares along each edge of the plate, then the unshaded area inthe centre of the plate is also a square.The area of this unshaded square in the centre of the plate is 1600 cm2,and so each of its sides has length

√1600 = 40 cm, as shown.

Consider the row of squares along the left edge of the plate.Since the side length of the square plate is 60 cm and the side length of theinner square is 40 cm, then the sum of the side lengths of the two shadedcorner squares is 60− 40 = 20 cm.Therefore, each shaded corner square (and thus each shaded square) hasside length 20

2= 10 cm.

60 cm 40 cm

(c) Using the same argument as in part (b), the area of the unshaded squarein the centre of the plate is 2500 cm2, and so each of its sides has length√

2500 = 50 cm, as shown.The side length of the square plate is 60 cm and so the sum of the sidelengths of 4 shaded squares (2 stacked vertically in the top two rows and2 stacked vertically in the bottom two rows) is 60− 50 = 10 cm.Therefore, each of these shaded squares (and thus each shaded square onthe plate) has side length 10

4= 5

2cm.

The side length of the square plate is 60 cm and each shaded squarehas length 5

2cm, and so along an outside edge of the plate there are

60÷ 52

= 60× 25

= 12× 2 = 24 shaded squares.There are 2 rows that each contain 24 shaded squares along each of thetop and bottom of the square, and 2 additional rows that each contain24− 4 = 20 shaded squares along the left and right sides of the square.That is, the total number of shaded squares on the plate is4× 24 + 4× 20 = 96 + 80 = 176.

60 cm 50 cm

3. (a) Triangle ABC is equilateral with side length 6, and so AB = BC = CA = 6.Since D is the midpoint of BC, then BD = DC = 3.In 4ADC, ∠ADC = 90◦ and so by the Pythagorean Theorem AD2 = AC2 −DC2.Therefore, h2 = 62 − 32 = 36 − 9 = 27 and so h =

√27 =

√9× 3 =

√9 ×√

3 = 3√

3,since h > 0.

2016 Galois Contest Solutions Page 4

(b) The shaded region lies inside the circle and outside the hexagon and thus its area isdetermined by subtracting the area of the hexagon from the area of the circle.First we find the area of the hexagon.Each vertex of hexagon EFGHIJ lies on the circle.Since the circle has centre O and radius 6, then OE = OF = OG = OH = OI = OJ = 6.Each side length of the hexagon is also 6, and so the hexagon is formed by six congruentequilateral triangles with side length 6. (For example, 4OGH is one these 6 triangles.)Each of these triangles is congruent to 4ABC from part (a) and thus has height h = 3

√3.

The area of each of the six congruent triangles is 12(6)(3

√3) = 9

√3.

Therefore, the area of hexagon EFGHIJ is 6× 9√

3 = 54√

3.The area of the circle with centre O and radius 6 is π(6)2 = 36π.Finally, the area of the shaded region is 36π − 54

√3.

(c) Let the area of the shaded region that we are required to find be A.Let the area of the shaded region in the diagram to the right be S.We may determine A by subtracting S from the area of the semi-circlewith centre P .First we determine S.Consider the circle with centre O. The shaded region having area Slies inside sector MON of this circle, but outside 4MON .

O

M NP

That is, S is determined by subtracting the area of4MON from the area of sector MON .In 4MON , MN = ON = OM = r (since ON and OM are radii), and so the triangle isequilateral. Join O to P .Since ON = OM and P is the midpoint of MN , then OP is the altitude (height) of4MON with base MN .In 4OPN , ∠OPN = 90◦ and so by the Pythagorean Theorem OP 2 = ON2 − PN2.

Since PN = 12(MN) = 1

2r, OP 2 = r2−(1

2r)2 = r2− 1

4r2 = 3

4r2, and so OP =

√34r2 =

√32r.

Therefore, the area of 4MON is 12(MN)(OP ) = 1

2(r)(√

32r)

=√34r2.

Next, we determine the area of sector MON .Since 4MON is equilateral, then ∠MON = 60◦.Thus, the area of sector MON is 60◦

360◦= 1

6of the area of the circle with centre O and

radius r, or 16πr2.

Therefore, S = 16πr2 −

√34r2.

Finally, one-half of the area of the circle with centre P and radius PN = 12r is

12π(12r)2

= 18πr2, and so A = 1

8πr2 − S = 1

8πr2 −

(16πr2 −

√34r2)

=(

18π − 1

6π +

√34

)r2.

Simplifying further, the exact area of the shaded region is 6√3−π24

r2.

4. (a) The prime factorization of 126 is 126 = 213271.Given an input of 126 = 213271, the output from the Barbeau Process is

126

(1

2+

2

3+

1

7

)= 126

(21 + 28 + 6

42

)= 3(55) = 165 .

(b) Given an input of p2q, the output from the Barbeau Process is p2q

(2

p+

1

q

)= 2pq + p2.

We are told that this output is equal to 135.Since 135 = 335, then 2pq + p2 = 335 or p(2q + p) = 335.

2016 Galois Contest Solutions Page 5

Since p is a prime number that is a divisor of 335, then p = 3 or p = 5.If p = 3, then 3(2q + 3) = 335 or 2q + 3 = 45 and so q = 21.However q is a prime number and 21 is not a prime number, so p 6= 3.If p = 5, then 5(2q + 5) = 335 or 2q + 5 = 27 and so q = 11.Therefore, the only pair (p, q) of different prime numbers that satisfies the given conditionsis (5, 11).

(c) Solution 1Given an input of 2a3b5c, the output from the Barbeau Process is

2a3b5c(a

2+b

3+c

5

)= 2a3b5c

(15a+ 10b+ 6c

30

).

We are told that this output is equal to 4× 2a3b5c.

Comparing these gives15a+ 10b+ 6c

30= 4 or 15a+ 10b+ 6c = 120.

Since a, b, c are positive integers, and 15× 8 = 120, 10× 12 = 120, and 6× 20 = 120, then1 ≤ a ≤ 7, 1 ≤ b ≤ 11, and 1 ≤ c ≤ 19.Further, 10b, 6c and 120 are divisible by 2, and so 15a = 120− 10b− 6c is divisible by 2.Therefore, a is divisible by 2 and since 1 ≤ a ≤ 7, then a equals 2, 4 or 6.Similarly, 15a, 6c, and 120 are divisible by 3, and so 10b is divisible by 3.Therefore, b is divisible by 3 and since 1 ≤ b ≤ 11, then b equals 3, 6 or 9.Finally, 15a, 10b, and 120 are divisible by 5, and so 6c is divisible by 5.Therefore, c is divisible by 5 and since 1 ≤ c ≤ 19, then c equals 5, 10 or 15.Since a ≥ 2, b ≥ 3, and c ≥ 5, then 15a ≥ 15 × 2 = 30, 10b ≥ 10 × 3 = 30, and6c ≥ 6× 5 = 30.That is, each of the terms 15a, 10b, 6c is at least 30, and so each of the terms is at most120− 2(30) = 60 (for example, 15a = 120− 10b− 6c ≤ 120− 30− 30 = 60).Therefore, 15a ≤ 60 and so a is equal to 2 or 4, 10b ≤ 60 and so b is equal to 3 or 6, and6c ≤ 60 and so c is equal to 5 or 10.If a = 2 and b = 3, then 6c = 120− 15(2)− 10(3) = 60, and so c = 10.If a = 2 and b = 6, then 6c = 120− 15(2)− 10(6) = 30, and so c = 5.If a = 4 and b = 3, then 6c = 120− 15(4)− 10(3) = 30, and so c = 5.If a = 4 and b = 6, then 6c = 120− 15(4)− 10(6) = 0, which is not possible.Therefore, the triples (a, b, c) of positive integers which satisfy the given conditions are(2, 3, 10), (2, 6, 5), and (4, 3, 5).

Solution 2Given an input of 2a3b5c, the output from the Barbeau Process is

2a3b5c(a

2+b

3+c

5

)= 2a3b5c

(15a+ 10b+ 6c

30

).

We are told that this output is equal to 4× 2a3b5c.

Comparing these gives15a+ 10b+ 6c

30= 4 or 15a+ 10b+ 6c = 120.

Since 10b, 6c and 120 are divisible by 2 and 15a = 120−10b−6c, then 15a is divisible by 2which means that a is divisible by 2. Thus, we set a = 2A for some positive integer A.Since 15a, 6c and 120 are divisible by 3 and 10b = 120−15a−6c, then 10b is divisible by 3which means that b is divisible by 3. Thus, we set b = 3B for some positive integer B.Since 15a, 10b and 120 are divisible by 5 and 6c = 120−15a−10b, then 6c is divisible by 5

2016 Galois Contest Solutions Page 6

which means that c is divisible by 5. Thus, we set c = 5C for some positive integer C.Therefore, the equation 15a + 10b + 6c = 120 becomes 30A + 30B + 30C = 120 orA+B + C = 4.Since A, B and C are positive integers that add to 4, the possible values for (A,B,C) are(2, 1, 1), (1, 2, 1), and (1, 1, 2), because each is at least 1 which only leaves 1 “left over” tomake up the total of 4.Since (a, b, c) = (2A, 3B, 5C), then the possible triples (a, b, c) are (4, 3, 5), (2, 6, 5), and(2, 3, 10).

(d) We proceed in a number of steps.

Step 1: Each exponent in the prime factorization must be a multiple of its primeWe show that, since the output is an integer multiple of the input, each exponent in theprime factorization must be an integer multiple of its corresponding prime.This generalizes what we found in Solution 2 to part (c).In other words, suppose that n = pa11 p

a22 p

a33 . . . pakk where p1, p2, ..., pk are different prime

numbers and a1, a2, ..., ak are integers that are each at least 0.

By definition, the output of the Barbeau Process is n

(a1p1

+a2p2

+a3p3

+ · · ·+ akpk

).

Note that we are allowing each of a1, a2, . . . , ak to be 0, which does not affect the value ofthe output.Here, we are told that the output of the Barbeau process is 3n.

Therefore, n

(a1p1

+a2p2

+a3p3

+ · · ·+ akpk

)= 3n or

a1p1

+a2p2

+a3p3

+ · · ·+ akpk

= 3.

Rearranging, we obtaina1p1

= 3− a2p2− a3p3− · · · − ak

pk.

Multiplying both sides by p2p3 · · · pk, we obtain

a1p2p3 · · · pkp1

= 3p2p3 · · · pk − a2p3 · · · pk − a3p2p4 · · · pk − · · · − akp2p3 · · · pk−1

Since every term on the right side is an integer, thena1p2p3 · · · pk

p1must be an integer.

Since the prime numbers p1, p2, . . . , pk are all distinct, then a1 must be a multiple of p1 tomake this fraction actually equal an integer.A similar argument will show that a2 is a multiple of p2, a3 is a multiple of p3, and so on.

Step 2: n does not have any prime factor larger than 7Suppose that n does have a prime factor that is larger than 7.In other words, suppose that n has a prime factor pi that is at least 11.In this case, the factor paii in the prime factorization of n is at least 1111 since pi ≥ 11 andai is a multiple of pi and so must also be at least 11.In this case, n ≥ 1111.But n is restricted to be less than 1010, so this is not possible.Therefore, n does not have any prime factor larger than 7.

Step 3: Simplifying the algebra

Combining Steps 1 and 2, n must be of the form n = 2a3b5c7d for some non-negativeintegers a, b, c, and d that are multiples of 2, 3, 5, and 7, respectively.As in Solution 2 to part (c), we set a = 2A, b = 3B, c = 5C, and d = 7D for somenon-negative integers A,B,C,D.

Therefore, the equationa1p1

+a2p2

+a3p3

+ · · ·+ akpk

= 3 becomesa

2+b

3+c

5+d

7= 3, which

2016 Galois Contest Solutions Page 7

becomes A+B + C +D = 3.Therefore, we want to find all of the non-negative integer solutions to A+B+C +D = 3which give n = 22A33B55C77D < 1010.

Step 4: Restrictions on A,B,C,DSince A+B +C +D = 3 and each of A,B,C,D is non-negative, then each of A,B,C,Dis at most 3.If D = 2 or D = 3, then n is divisible by 714 or 721, each of which is larger than 1010.Therefore, D ≤ 1.If C = 3, then n is divisible by 515, which is larger than 1010.Therefore, C ≤ 2.

Step 5: Determining the values of nSince A + B + C + D = 3 and each of A,B,C,D is non-negative, then A,B,C,D couldbe (i) 3, 0, 0, 0 in some order, or (ii) 2, 1, 0, 0 in some order, or (iii) 1, 1, 1, 0, in some order.(If one value is 3, the rest must be 0. If one value is 2, then there must be one 1 and two0s. If no value is 2, then there must be three 1s and one 0.)Using the facts that C ≤ 2 and D ≤ 1, we enumerate the possibilities:

Category A B C D n Less than 1010?(i) 3 0 0 0 26 Yes(i) 0 3 0 0 39 Yes(ii) 2 1 0 0 2433 Yes(ii) 2 0 1 0 2455 Yes(ii) 2 0 0 1 2477 Yes(ii) 1 2 0 0 2236 Yes(ii) 0 2 1 0 3655 Yes(ii) 0 2 0 1 3677 Yes(ii) 1 0 2 0 22510 Yes(ii) 0 1 2 0 33510 Yes(ii) 0 0 2 1 51077 No(iii) 1 1 1 0 223355 Yes(iii) 1 1 0 1 223377 Yes(iii) 1 0 1 1 225577 No(iii) 0 1 1 1 335577 No

In each case, we can check whether the possible value of n is smaller than 1010 usinga calculator. (Which of these calculations can you “reason” through without using acalculator?)

In summary, the possible values of n are

26, 39, 2433, 2455, 2477, 2236, 3655, 3677, 22510, 33510, 223355, 223377